A274743 Repunits with odd indices multiplied by 99, i.e., 99*(1, 111, 11111, 1111111, ...).

99, 10989, 1099989, 109999989, 10999999989, 1099999999989, 109999999999989, 10999999999999989, 1099999999999999989, 109999999999999999989, 10999999999999999999989, 1099999999999999999999989, 109999999999999999999999989, 10999999999999999999999999989

Offset: 1

Author

Rodolfo A. Fiorini, Jul 04 2016

Keywords

• nonn
• easy

Comments

It is apparent that the reciprocals of the terms in the sequence give an increasing sequence of periodic terms similar to A095372, but with the initial term equal to "01". The leading zero is important (see links). Furthermore, the reciprocals of the terms give a sequence of even growing periods, starting from 2, with delta = 4 (i.e., 2, 6, 10, 14, 18, ...).

Adding "11" to each term gives the binary representation of the n-th iteration of "Rule 14" elementary cellular automaton starting with a single ON (black cell) as in A266299.

Examples

			a(2) = 101*10989 - 100*99 = 1099989.
		

Links

• Rodolfo A. Fiorini, How Random is Your Tomographic Noise? A Number Theoretic Transform (NTT) Approach, Fundamenta Informaticae, 135(1-2), 2014, 135-170.
• Rodolfo A. Fiorini, Computerized tomography noise reduction by CICT optimized exponential cyclic sequences (OECS) co-domain, Fundamenta Informaticae, vol.141 (2015), 115-134.
• Index entries for linear recurrences with constant coefficients, signature (101,-100).

Crossrefs

Cf. A002275, A095372, A100706, A266299, A274755.

Programs

Magma

[11*(10^(2*n-1)-1) : n in [1..20]]; // Wesley Ivan Hurt, Jul 04 2016

Maple

A274743:=n->11*(10^(2*n-1)-1): seq(A274743(n), n=1..20); # Wesley Ivan Hurt, Jul 04 2016

Mathematica

Array[99(10^(2 # - 1) - 1)/9 &, 15] (* Michael De Vlieger, Jul 04 2016 *)
99*Table[FromDigits[PadRight[{},2n+1,1]],{n,0,15}] (* Harvey P. Dale, Jul 22 2019 *)

PARI

Vec(99*x*(1+10*x)/((1-x)*(1-100*x)) + O(x^99)) \\ Altug Alkan, Jul 05 2016

PARI

a002275(n) = (10^n-1)/9
a(n) = 99*a002275(2*n-1) \\ Felix Fröhlich, Jul 05 2016

Formula

a(n) = 101*a(n-1) - 100*a(n-2) for n>2, with a(0)= 99 and a(1)= 10989.

a(n) = 99*A100706(n-1).

G.f.: 99*x*(1 + 10*x)/((1 - x)*(1 - 100*x)). - Ilya Gutkovskiy, Jul 04 2016

a(n) = 11*(10^(2*n-1)-1). - Wesley Ivan Hurt, Jul 04 2016

E.g.f.: 11*(9 - 10*exp(x) + exp(100*x))/10. - Stefano Spezia, Aug 05 2024